Well, one can note that $142857$ is a bit of a special number in that $\frac{1}{7}=0.\overline{142857}$, which means that, by long division, one obtains $\overline{142857}=142857142857142857...$ ad ...

The highest power of $2$ dividing $n!$ is $\lfloor \frac{n}{2} \rfloor+\lfloor \frac{n}{4} \rfloor+\lfloor \frac{n}{8} \rfloor+...$ The highest power of $5$ dividing $n!$ is $\lfloor \frac{n}{5} \... View answer 7 votes Another possible solution: just compute the geometric sum in the exponent. $$x=6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...}=6$$ Could be done in your head if you wanted to. View answer Accepted answer 6 votes Here's a very simple answer for your second question, containing some essential rules of thumb to have in mind to understand basic ring theory, and in particular that of PIDs such as$\mathbb{Z}$. In ... View answer Accepted answer 5 votes The usual way to go about this classic problem is the following. Let$G=\cup G_d$where$G_d$is the set of elements of$G$of order$d$for each$d|n$. Since there's at most one subgroup of order$d$,... View answer Accepted answer 5 votes Hint :$\sum a_k\times10^k\equiv\sum a_k \pmod {9}$View answer 4 votes A simple algorithm is the following : Take integer part$\rightarrow$multiply the numerator of the remainder by$10$(in base$10$)$\rightarrow$divide by the denominator$\rightarrow$repeat ... View answer Accepted answer 3 votes Use a contour integral in the upper complex plane if you know residue calculus.$\int_0^{+\infty}\frac{x\sin(x)}{1+x^2}dx=\frac12 \operatorname{Im} \int_{\mathbb{R}}\frac{xe^{ix}}{1+x^2}=\frac 12\...

$2^{29}\equiv2^{-1}\equiv\frac{1}{2}\equiv\frac{10}{2}\equiv5\equiv \text{s} \pmod 9$ where $\text{s}$ denotes the sum of its digits. The sum of the ten digits is $\text{S}=0+1+2+3+4+5+6+7+8+9=\frac{... View answer 3 votes$1^2\equiv4^2\equiv1 \pmod 51\times4\equiv4\times1\equiv4 \pmod 5$So yes, the remainder of the product is always either$1$or$4$View answer Accepted answer 3 votes $$a^{(\log_ab)^2}=a^{\log_a(b)\times \log_a(b)}=b^{\log_a(b)}$$ View answer 3 votes Draw a tree and notice that, by periodicity, the probability that B wins is : $$P_B=\frac{5}{6}+\frac{1}{6} \times \frac{1}{6} \times(\frac{5}{6}+...)$$ So we can express this probability as the ... View answer 2 votes If$p \leq n$is prime then it has exponent$\lfloor \frac{n}{p} \rfloor+\lfloor \frac{n}{p^2}\rfloor +\cdots+\lfloor \frac{n}{p^\alpha}\rfloor$in$n!$where$\alpha$is the largest integer$\geq 1$... View answer Accepted answer 2 votes If$A_t$and$B_t$are independent Brownian motions, then$\sin(a)A_t+\cos(a)B_t$is a centered Gaussian process (due to independence,$\sin(a)A_t+\cos(a)B_t$has law$N(0,(\sin(a)^2+\cos(a)^2)t)=N(0,...

We have that $P^2=578+2\sqrt{(x-144)(722-x)}$ To maximize $P^2$ is to maximize $(x-144)(722-x)$, which is a downward parabola. By symmetry, it is maximized right between the two roots, i.e. at $x=\... View answer 2 votes$\int_0^{+\infty}xe^{-ax^2}dx=\frac{1}{-2a}\int_0^{+\infty}-2axe^{-ax^2}dx=\frac{1}{-2a}[e^{-ax^2}]_0^{+\infty}=\frac{1}{2a}$View answer 2 votes Answering self:$\sigma_2>\sigma_1$The two problems are not equivalent because in random walk$1$, a random path is chosen uniformly among the possible ones, but not in random walk$2$Indeed, ... View answer 2 votes$65=13\times5870^{479}\equiv0 \pmod{5}\color{red}{87}0^{479}\equiv\color{red}{9}0^{479}\equiv(-1)^{479}\equiv-1\pmod{13}$So$x=870^{479}=-1+13k\equiv0 \pmod 5 \rightarrow k\equiv\frac{1}{13}\...

I can't find an easy way to show this by hand (although it is still possible). However, I am presenting an answer assuming the prime factorization of $z$ is known (or that the use of a calculator is ...

$(m+n)^2=m^2+n^2+2mn\equiv m^2+n^2 \pmod 2$ So $m^2+n^2$ odd implies $(m+n)^2$ odd.

I think I fixed the proof, and it is rather elementary, but not so simple... We're working with $N>1$ ($N=1=1^2+2\times0^2$ has no inexpressible prime factor) Notice that according to ...

Let's work $\pmod {100}$ Notice that $43^2=(50-7)^2\equiv7^2\equiv49 \pmod {100}$ and likewise $49^2\equiv1 \pmod {100}$ Therefore, $43^4\equiv 1 \pmod {100}$ $23^{33}\equiv (-1)^{33}\equiv-1 \... View answer Accepted answer 2 votes $$\sum_{i=1}^n i3^i=3+2\times3^2+3\times3^3+...+n\times3^n=(3+3^2+...+3^n)+(3^2+3^3+...+3^n)+...+3^n=\sum_{i=1}^n \sum_{k=i}^n3^k=\sum_{i=1}^n\frac{3^i-3^{n+1}}{1-3}=\frac{1}{2}\sum_{i=1}^n(3^{n+1}-3^... View answer 2 votes$$\sum_{n=0}^{+\infty}2^{-3n}=\sum_{n=0}^{+\infty}(2^{-3})^n=\sum_{n=0}^{+\infty}(\frac{1}{8})^n=\frac{1}{1-\frac{1}{8}}=\frac{1}{\frac{7}{8}}=\frac{8}{7}$$View answer Accepted answer 2 votes 100=4\times25 and gcd(4,25)=1 14^{200}\equiv2^{200}\times7^{200}\equiv0 mod 4 (since 4=2^2) 14^{200}\equiv(-11)^{200}\equiv11^{200}\equiv121^{100}\equiv(-4)^{100}\equiv4^{100}\equiv(-1)^{... View answer 2 votes U=(N_{black},N_{white}) At the start : U_1=U_2=U_3=(5,7) The probability of picking a random black ball from U_1 is \frac{5}{12}, the probability of picking a white one is \frac{7}{12} ... View answer 2 votes 2016 = 365 \times 5+191 So 5 years and 189 days have elapsed (191-2 days because there are two leap year, one every 4 years, except when the year number is divisible by 4, and 100 but not by 400). ... View answer 1 votes The number of ways is$$\binom{4}{2}\binom{7}{4}+\binom{4}{3}\binom{7}{3}+\binom{4}{4}\binom{7}{2} =6\times35+4\times35+21 = 371$$"At least two women" means either$2$,$3$or$4$. View answer Accepted answer 1 votes First, pick two friends, this yields$\binom{3}{2}$. Then arrange them, this yields a factor of$2!\$. Then, note that the "exactly" means the third not-picked friend should not sit next to ...